Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int fa[100005][18], dep[100005];
vector<int> E[100005], cit[100005];
struct node {
  int sz, s[15];
  node() {
    sz = 0;
    memset(s, 0, sizeof(s));
  }
  node operator+(const node &A) {
    node ret;
    int i = 0, j = 0, k = 0;
    while (i < sz && j < A.sz) {
      if (k >= 12) break;
      if (s[i] < A.s[j])
        ret.s[k++] = s[i++];
      else
        ret.s[k++] = A.s[j++];
    }
    while (k < 12 && i < sz) ret.s[k++] = s[i++];
    while (k < 12 && j < A.sz) ret.s[k++] = A.s[j++];
    ret.sz = k;
    return ret;
  }
  void operator+=(const vector<int> &A) {
    int tmp[15];
    int i = 0, j = 0, k = 0;
    while (i < sz && j < A.size()) {
      if (k >= 12) break;
      if (s[i] < A[j])
        tmp[k++] = s[i++];
      else
        tmp[k++] = A[j++];
    }
    while (k < 12 && i < sz) tmp[k++] = s[i++];
    while (k < 12 && j < A.size()) tmp[k++] = A[j++];
    sz = k;
    for (int i = 0; i < sz; i++) s[i] = tmp[i];
  }
} Po[100005][18];
void Print(node a) {
  printf("%d", a.sz);
  for (int i = 0; i < a.sz; i++) printf(" %d", a.s[i]);
  puts("");
}
void dfs(int x, int f, int d) {
  fa[x][0] = f;
  Po[x][0] += cit[x];
  dep[x] = d;
  for (int i = 0; i < E[x].size(); i++) {
    int v = E[x][i];
    if (v == f) continue;
    dfs(v, x, d + 1);
  }
}
void Up(node &key, int &x, int step) {
  for (int i = 0; i <= 17; i++)
    if (step & (1 << i)) {
      key = key + Po[x][i];
      x = fa[x][i];
    }
}
node LCA(int a, int b) {
  node key;
  if (dep[a] > dep[b]) swap(a, b);
  Up(key, b, dep[b] - dep[a]);
  if (a == b) {
    key = key + Po[a][0];
    return key;
  }
  for (int i = 17; i >= 0; i--)
    if (fa[a][i] != fa[b][i]) {
      key = key + Po[a][i];
      a = fa[a][i];
      key = key + Po[b][i];
      b = fa[b][i];
    }
  key = key + Po[a][0] + Po[b][0] + Po[fa[a][0]][0];
  return key;
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    E[a].push_back(b);
    E[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    int a;
    scanf("%d", &a);
    cit[a].push_back(i);
  }
  dfs(1, 0, 1);
  for (int i = 1; i <= 17; i++)
    for (int j = 1; j <= n; j++) {
      fa[j][i] = fa[fa[j][i - 1]][i - 1];
      Po[j][i] = Po[j][i - 1] + Po[fa[j][i - 1]][i - 1];
    }
  for (int i = 1; i <= q; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    node z = LCA(a, b);
    z.sz = min(z.sz, c);
    Print(z);
  }
  return 0;
}